Question

what is the mean and variance of samples of three cards are drawn random from a population of seven cards numbered 1 to 7

Answer 1

We have population values 1,2,3,4,5,6,7 population size N=7 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+2+3+4+5+6+7}{7}=4$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{9+4+1+0+1+4=9}{7}=4$$ $$\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.4142$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{7}C_3=35.$

Mean of sampling distribution 

$$\mu_{\bar{X}}=\mu=4$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$= \dfrac{4}{3}(\dfrac{7-3}{7-1})=\dfrac{8}{9}$$

The number of possible samples which can be drawn with replacement is $N^n=7^3=343.$

Mean of sampling distribution 

$$\mu_{\bar{X}}=\mu=4$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}= \dfrac{\sigma^2}{n}= \dfrac{4}{3}$$